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yuradex [85]
3 years ago
5

If an input force of 202N is applied to the handles of the wheelbarrow in the sample problem how large is the output force that

just lifts the load?
Physics
1 answer:
Sloan [31]3 years ago
3 0

For this case, the relationship between the two forces is given by:

F1 = nF2

Where,

  • F1: output strength
  • F2: input force
  • n: mechanical advantage

Then, replacing the values we have:

F1 = (2.2)(202)

Having the calculations we have:

F1 = 444.4N

Answer: The output force that only lifts the load is F1 = 444.4N.

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A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move
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W=-2.1405\times 10^9\,J

Explanation:

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electric field, E=148\,V.m^{-1}

charge, Q=-13\,\mu C=-13\times 10^{-6}\,C

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final position coordinates,   p2 =(107,76)

We find the distance through which the charge has been moved:

d=\sqrt{(x1-x2)^2+(y1-y2)^2}

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

d=\sqrt{(107-(-81))^2+(76-(-131))^2}

d= 279.63\,m

Now we need the angle through which displacement is made with respect to the direction of electric field.

tan\,\theta= \frac{y2-y1}{x2-x1}

\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]

\theta= 47.75^{\circ}

Now from the relation between the change in potential difference:

\Delta V= E.d.cos\,\theta

\Delta V= 148\times 279.63\times cos\,47.75^{\circ}

\Delta V= 27826.06 V

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∴\Delta V=\frac{W}{Q}

W=\frac{\Delta V}{Q}

Putting the respective values

W=\frac{27826.06 }{-13\times 10^{-6}}

W=-2.1405\times 10^9\,J

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3 years ago
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